1. Field of the Invention
The invention relates to an active double-sided rectifier circuit which is particularly suitable for use with a current transformer which is used, for example, in an earth leakage detector. Such detectors must be able to indicate very weak currents and, because they are installed in distributing boxes, must not have large dimensions. However, the invention is obviously not restricted to this use.
2. Description of the Related Art
The most usual circuit for rectifying a current from a current transformer is formed by a diode bridge composed of four diodes and connected to the secondary winding of the current transformer. This rectifier circuit has the advantage of full-wave rectification and consists of few, passive components, while the output is potential free, so that there is free choice in respect of direct current level.
This known circuit nevertheless has disadvantages making it less suitable, for example, for earth leakage detectors. When the diode bridge output is loaded, this loading will adversely affect the transmission of the current transformer. This transmission is ideal only with short-circuited output terminals. The voltage drop across the diodes in the forward direction also influences the transmission. This voltage drop may be of the order of 600 mV in the case of of an Si diode.
FIGS. 1 and 2 show equivalent circuit diagrams of current transformers. In FIG. 2, L is the self-induction of the secondary windings and R the output resistance.
For the output voltage U.sub.o : ##EQU1##
Transmission becomes maximum at R=.infin., when ##EQU2##
A small toroidal core transformer having 1000 turns has a self-inductance of about 5H. When the primary current to be measured amounts to 20 mA, and with a mains frequency of 50 Hz the output voltage U.sub.o is: EQU U.sub.o =2.pi..50.5.20 10.sup.6 =31 mV (3)
Because of the abovementioned diode voltage of 600 mV in the forward direction, the low voltage of 31 mV mentioned cannot be rectified with a passive diode bridge of this kind.
This shortcoming can be avoided by making the current transformer heavier, with a secondary winding with high self-inductance. This, however, leads to an undesired larger volume and to a higher price.
These disadvantages can be obviated by using a rectifier circuit having active components, such as operational amplifiers and transistors. The price and the volume of these components justify their use, and in addition they can be made in the form of an integrated unit, while use may be made of a small and inexpensive current transformer.
FIG. 3 shows an example of a possible circuit with active components and half-wave rectification.
The current transformer with a transmission ratio of 1:n is provided with a primary winding 1 and a secondary winding 2. A connection terminal of the secondary winding is connected via a capacitor 3 or a resistor 9, shown above it, to an input terminal of the operational amplifier 4, while the other connection terminal of the secondary winding 2 is connected to the other input terminal of the operational amplifier 4. The output of the operational amplifier 4 is connected to the base of the transistor 6 and likewise to the anode of the diode 5. The emitter of the transistor 6 and the cathode of the diode 5 are connected to the negative input of the operational amplifier. Because the input impedance of the feedback amplifier is very low, the current transformation will be ideal, namely I.sub.2 =1/n I.sub.1.
The output voltage U.sub.o occurs through the output resistor 7. The transistor circuit is supplied by a current source 8. In this half-wave rectifier circuit, the following applies to the positive half-cycle of a sinusoidal alternating current: EQU U.sub.o =I.sub.2 .multidot..alpha..multidot.R.sub.L. (5)
I.sub.2 is here the current through the secondary winding, .alpha. is about 0.99, that is to say the ratio between the collector current and the emitter current of the transistor 6, while R.sub.L represents the value of the output resistor 7.
During the negative half-cycle the output voltage U.sub.o is equal to 0. The following therefore applies to the mean value U.sub.o when n is for example selected at 1000: ##EQU3##
Here U.sub.o and I.sub.o represent the mean values.
However, this circuit once again has some disadvantages. In order to prevent the input offset of the operational amplifier from causing heavy currents to flow through the secondary winding and, through the action of the operational amplifier, also through the diode and the transistor, this voltage link to the secondary winding for direct current would have to be interrupted; in FIG. 3 this is done with the aid of the capacitor 3. In order however to allow sufficient current to pass with a mains frequency of 50 Hz, the capacitance of this capacitor 3 would have to be fairly high, at least 5 .mu.F, which is possible only with two electrolytic capacitors connected oppositely in series. This, however, is an expensive solution.
This expensive solution can be avoided by using, instead of a capacitor 3, the resistor 9, shown above the capacitor 3, in the input connection to the secondary winding. This input resistor 9 must then have a sufficiently high ohmic value so that the current produced by the offset will in all cases be kept sufficiently low.
If for this input resistor 9 a value of 1 k .OMEGA. is selected and if a value of 1 M .OMEGA. is selected for the output resistor 7, then for an offset of 2 mV an output voltage through the resistor 7 of 2 V is obtained, namely: ##EQU4##
This 2 V output offset will be too high for almost all applications and thus cannot be accepted.
The use of only one half-cycle because of the single-sided rectification also leads to additional time delay and to ripple.